I have attempted to give some quick answers to your issues. The best way of seeing what is going on is to carry out lots of substitutions of solutions into DEs and especially see how using the product rule generates the extra terms in the original DE.
1. What is an "exact" differential equation.
Is it just a differential equation that can be solved using simple methods?
Well kind of, the point is that if you just wrote down some functions of x and y and differential with an equals sign, there might not be a function that you can plug in that would work. In fact, if you got a computer program and randomly generated DEs, most of them would be insoluble.
So an exact DE is just one for which a solution exists which is actually a function.
Edit
I neglected to mention an important thing that is often implied but not stated about the solution to Homogenous Equations.
If f(x,y) and g(x,y) add up to zero (which is the kind of thing you will get when you substitute solutions into your original HE) then unless they are trivial or the same function then they must independently be zero. This is an important bit of insight to helping you understand the answer to 4 above.
2. For a second order differential equation in the form:
ay'' + by' + cy = 0
With exact roots, why is the solution y= (A+Bx)e^Cx
I understand the solution for non equal roots as it is given in the textbook
Your teacher is correct here, for a lot of differential equations, a long time ago someone clever (like Euler maybe) just sat down with a piece of paper and tried different functions and given experience of how the exponential function works
they just saw whether particular function could possibly work.
In your case why not try a few such as some trig functions?
3. For a second order differential equation in the form: ay'' + by' + cy = f(x)Why is the solution y = complementary function + particular integral
How does the particular integral compensate for the addition of f(x).
I though all that would be required would be to integrate the right hand side with respect to x twice.
If you could find two solutions of the nonhomogeneous equation (I call this NH) (i.e. f(x)=not 0) then their difference would be a solution of the homogeneous equation (f(x)=0) (I will call the equ the HE)
Lets imagine the are NH1 and NH2 sop we get NH1-NH2 = H1 hence NH1 = H1 + NH2 which is what the CF and PI are.
4. Why in questions of the above form, if the particular integral has a term which is part of the complementary function, the particular integral is not correct.
Why is this?
Because if you have a CF (a solution of the HE) which is two terms added together then any constant multiplier of either function will also be a solution of the HE. The constant can be 1. So the PI you have is a solution of the HE and can't therefore be part of the solution to the NH
5. Why you then able to multiply the value of the integral by a power of x (assuming that power isn't in the complementary function. Would you be able to do this with a second differential equation where you didn't need to this and would it still work?
The best I can give you for this is that when you substitute the x times PI into the original NH DE, you will get extra terms which means that the CF + PI is no longer a solution to the HE because the LHS is no longer adding up to zero.